Question 1: The sum of two numbers is 60 and their difference is 14 . Find the numbers.

Answer:

Let the two numbers be x \ and \ y

 x+y=60  ... ... ... ... ... ... (i)

  x-y=14 ... ... ... ... ... ... (ii)

Add  (i) and (ii)

2x=74 \ or\  x=37

Therefore y=37-14=23

Hence the two numbers are 23 \ and\  37

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Question 2: Twice a number is equal to thrice the other number. If the sum of the numbers is 85 , find the numbers.

Answer:

Let the two numbers are x \ and \ y

2x=3y  ... ... ... ... ... ... (i)

x+y=85 ... ... ... ... ... ... (ii) 

Solving for x \ and\  y 

 2x-3y=0 

\underline {(-)2x+2y=170 }

-5y=-170 

or y=34 

Therefore x= \frac{3}{2} \times 34=51

Hence the two numbers are 51 \ and\  34 .

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Question 3: Find two numbers such that twice the first added to thrice the second gives 70 and twice the second added to thrice the first gives 75 .

Answer:

Let the numbers be x \ and \ y

2x+3y=70 ... ... ... ... ... ... (i)

3x+2y=75 ... ... ... ... ... ... (ii)

Solving for x \ and \ y

Multiply i) by 2 and ii) by 3 and subtract ii) from i)

 6x+9y=210

\underline{(-)  6x+4y=150}

5y=60

\Rightarrow y=12

Substituting in i)

\Rightarrow x= \frac{70-3y}{2} = \frac{70-36}{2} =17

Hence the two numbers are 17 \ and\  12

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Question 4: Find two numbers which differ by 9 and are such that four times the larger added to three times the smaller gives 92 .

Answer:

Let the numbers be x \ and \ y

x-y=9  ... ... ... ... ... ... (i)

4x+3y=92 ... ... ... ... ... ... (ii)

Solving by x \ and\  y

Multiply i) by 4 and subtract ii) from i)

  4x-4y=36

\underline{(-)4x+3y=92}

-7y=-56

\Rightarrow y=8

Calculating for x

\Rightarrow x=y+9=8+9=17

Hence the two numbers are 8 \ and \  17

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Question 5: The sum of two numbers is 30 and the difference of their squares is 180 . Find the numbers.

Answer:

Let the two numbers be x \ and \ y

x+y=30  ... ... ... ... ... ... (i)

x^2-y^2=180 ... ... ... ... ... ... (ii)

Simplifying ii)

(x-y)(x+y)=180

\Rightarrow x-y=180/30=6

x-y=6 ... ... ... ... ... ... (iii)

Solving for x \ and\  y

Add i) and iii)

 x+y=30

\underline{(+)   x-y=6} 

2x =36

\Rightarrow x=18

Substituting in i) y=30-18=12

Hence the numbers are 18 \ and \  12

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Question 6: The sum of the digits of a two-digit number is 8 . On adding 18 to the number, its digits are reversed. Find the number.

Answer:

Let the two digit numbers be xy

x+y=8  ... ... ... ... ... ... (i)

10x+y+18=10y+x ... ... ... ... ... ... (ii)

Simplifying  ii)

9x-9y=-18

\Rightarrow x-y=-2 ... ... ... ... ... ... (iii)

Solving for x \ and\  y

Add i) and iii)

x+y=8

\underline{(+)    x-y=-2}

2x=6

\Rightarrow x=3

Substituting in i)  \Rightarrow y=8-3=5

Hence the numbers is 35

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Question 7: Two digit number is three times the sum of its digits. lf 45 is added to the number, its digits are reversed. Find the original number.

Answer:

Let the two digit numbers be xy

3(x+y)=10x+y  ... ... ... ... ... ... (i)

10x+y+45=10y+x ... ... ... ... ... ... (ii)

Simplifying  i) and ii)

7x-2y=0 ... ... ... ... ... ... (iii)

9x-9y=-45

\Rightarrow x-y=-5 ... ... ... ... ... ... (iv)

Solving  for x \ and\  y

Multiplying iv) by 7 and Subtract iv) from ii)

 7x-2y=0

\underline{(-)7x-7y=-35}

5y=35

\Rightarrow y=7

Hence x=y-5=7-5=2

Therefore the numbers is 27

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Question 8: A two-digit number is seven times the sum of its digits. If 27 is subtracted from the number, its digits get interchanged. Find the number.

Answer:

Let the two digit number be xy

10x+y=7(x+y)

\Rightarrow 3x-6y=0  ... ... ... ... ... ... (i)

10x+y-27=10y+x

\Rightarrow 9x-9y=27

\Rightarrow x-y=3 ... ... ... ... ... ... (ii)

Solving for x  \ and\  y

Multiplying ii) by 3 and Subtract ii) from i)

3x-6y=0

(-)   3x-3y=9

-3y=-9 

\Rightarrow y=3

Hence x=y+3=6

Therefore the numbers is 63

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Question 9: Find a fraction which reduces to 2/3 when 3 is added to both its numerator and denominator; and reduces to 3/5 when 1 is added to both its numerator and denominator.

Answer:

Let the fraction be \frac{x}{y} 

\frac{x+3}{y+3} = \frac{2}{3}

\Rightarrow 3x+9=2y+6

\Rightarrow 3x-2y=-3  ... ... ... ... ... ... (i)

(x+1)/(y+1)=3/5

\Rightarrow 5x+5=3y+3

\Rightarrow 5x-3y=2 ... ... ... ... ... ... (ii)

Solving for x \ and \ y

Multiplying i) by 5 and ii) by 3 and subtract ii) from i)

15x-10y=-15

\underline {(-)15x-9y=-6}

-y=-9

\Rightarrow y=9

Hence x= \frac{2y-3}{3} = \frac{18-3}{3} =5

Therefore the fraction is  \frac{5}{9} 

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Question 10: On adding 1 to the numerator of a fraction, it becomes 1/2 . Also, on adding 1 to the denominator of the original fraction, it becomes 1/3 . Find the original fraction.

Answer:

Let the fraction be \frac{x}{y} 

\frac{x+1}{y} = \frac{1}{2}   

\Rightarrow 2x-y=-2  ... ... ... ... ... ... (i) 

\frac{x}{y+1} = \frac{1}{3}   

\Rightarrow 3x-y=1 ... ... ... ... ... ... (ii) 

Solving for x \ and \ y

Multiplying i) by 3 and ii) by 2 and subtract ii) from i)

   6x-3y=-6 

\underline{ (-)   8x-2y=2} 

-y=-8 

\Rightarrow y=8 

Hence x= \frac{1+8}{3} =3 

Therefore the fraction is  \frac{3}{8} 

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Question 11: In a given fraction, if the numerator is multiplied by 2 and the denominator is reduced by 5 , we get 6/5 . But, if the numerator of the given fraction is increased by 8 and the denominator is doubled, we get 2/5 . Find the fraction.

Answer:

Let the fraction be \frac{x}{y} 

 \frac{2x}{y-5} = \frac{6}{5}    

 \Rightarrow 10x-6y=-30  ... ... ... ... ... ... (i) 

 \frac{x+8}{2y} = \frac{2}{5}    

 \Rightarrow 5x-4y=-40 ... ... ... ... ... ... (ii) 

Solving for x \ and \ y

Multiplying ii) by 2 and subtract ii) from i

 10x-6y=-3 

 \underline{(-)10x-8y=-80} 

 2y=50   

 \Rightarrow y=25 

Hence  x=(4y-40)/5=(100-40)/5=12 

Therefore the fraction in   \frac{12}{25}   

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Question 12: 5 years ago, a lady was thrice as old as her daughter. 10 years hence, the lady would be twice as old as her daughter. What are their present ages?

Answer:

Let the present age of lady  =x

Let the Present age of Daughter  = y

 x-5=3(y-5)  

 \Rightarrow x-3y=-10  ... ... ... ... ... ... (i)

 x+10=2(y+10)  

 \Rightarrow x-2y=10 ... ... ... ... ... ... (ii)

Solving for x \ and \ y

Subtract ii) from i)

 x-3y=-10

 \underline{(-)   x-2y=10}

 -y=-20  

 \Rightarrow y=20

Hence  x=2y+10=50

Therefore :

Lady’s Age  = 50 \ years

Daughter’s Age  = 20 \ years

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Question 13: The sum of the ages of A and B is 39 years. In 15 years’ time, the age of A will be twice the age of B. Find their present ages.

Answer:

Let A’s Age =x

Let B’s Age =y

x+y=39  ... ... ... ... ... ... (i)

x+15=2(y+15)  

\Rightarrow x-2y=15 ... ... ... ... ... ... (ii)

Solving for x \ and \ y

Subtract ii) from i)

  x+y=39

\underline{(-)  x-2y=15}

3y=24

or y=8

\Rightarrow x= 39-y=39-8=31

Therefore

A’s Age =31 \ years

B’s Age =8\  years

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Question 14: A is 15 years elder than B. 5 years ago A was four times as old as B. Find their present ages.

Answer:

Let the age of B =x,

\Rightarrow Age \ of \ A=x+15

+15-5=4(x-5)

\Rightarrow x+10=4x-20

3x=30 \ or\  x=10

Hence A’s Age = 10+15=25

Therefore

B’s Age =10 \ years

A’s Age =25  \ years

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Question 15: Six years ago, the ages of Geeta and Seema were in the ratio 3 : 4 . Nine years hence, their ages will be in the ratio 6 : 7 . Find their present ages.

Answer:

Let the age of Geeta =x

Let the age of Seema =y

\frac{x-6}{y-6} = \frac{3}{4}

  \Rightarrow 4x-24=3y-18

    \Rightarrow 4x-3y=6  ... ... ... ... ... ... (i)

\frac{x+9}{y+9} = \frac{6}{7}

      \Rightarrow 7x+63=6y+54

7x-6y=-9 ... ... ... ... ... ... (ii)

Solving for x \ and \ y

Multiplying i) by 7 and ii) by 4 and Subtract ii) from i)

28x-21y=42

\underline{(-)28x-24y=-36}

3y=78  

\Rightarrow y=26

Hence x= \frac{6+3\times 26}{4} =21

Therefore

Geeta’s Age =21 \ years

Seema’s Age =26 \ years

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Question 16: 4 knives and 6 forks cost Rs. 200 , while 6 knives and, 7 forks together cost Rs. 264 . Find the cost of a knife and that of a fork.

Answer:

Let the cost of knives = x

Let the Cost of fork = y

4x+6y=200  ... ... ... ... ... ... (i)

6x+7y=264 ... ... ... ... ... ... (ii)

Solving for x \ and \ y

Multiplying i) by 3 and ii) 2

12x+18y=600

\underline{(-)12x+14y=528}

4y=72

\Rightarrow y=18

Hence x= \frac{200-6\times 18}{4} =23

Hence Cost of :

Knive =23 \ Rs.

Fork =18 \ Rs.

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Question 17: The cost of 13 cups and 16 spoons is Rs. 296 , while the cost of 16 cups and 13 spoons is Rs. 284 . Find the cost of 2 cups and 5 spoons.

Answer:

Let the cost of Cup = x

Let the Cost of Spoon =y

13x+16y=296  ... ... ... ... ... ... (i)

16x+13y=284 ... ... ... ... ... ... (ii)

Solve for x \ and \ y

Multiply i) by 16 and ii) by 13 and subtract ii) from i)

208x+256y=4736

\underline{(-)     208x+169y=3692}

87y=1044

\Rightarrow y=12

Hence x= \frac{296-16\times 12}{13} =8

Hence Cost of:

Cup =Rs.\ 8

Spoon =Rs.\  12

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Question 18: Rahul covered a distance of 128 km in 5 hours, partly on bicycle at 16 kmph and partly on moped at 32 kmph. How much distance did he cover on moped?

Answer:

Total Distance =128 \ km

Total Time =5 \ Hrs.

Let the distance covered by cycle =x \ km

Let the Distance covered by moped = (120-x) \ Km.

\Rightarrow     \frac{x}{16} + \frac{120x}{32} =5

Simplify \Rightarrow 2x+128-x=160

\Rightarrow x=160-128=32

Hence Distance Covered by:

Cycle =32\ Km.

Moped = 96\ Km.

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Question 19: A boat can go 75 km downstream in 5 hours and, 44 km upstream in 4 hours. Find i) the speed of the boat in still water (ii) the rate of the current.

Answer:

Let the Speed of boat in still water = x

Speed of Stream = y

75/(x+y)=5 

\Rightarrow x+y=17 ... ... ... ... ... ... (i)

4x/(x-y)=4

\Rightarrow x-y=11 ... ... ... ... ... ... (ii)

Solve for x \ and \ y ,  add i) and ii)

2x=28 or x=14 \ Km /hr

Speed of Stream =3 \ Km/hr

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Question 20: The monthly incomes of A and B are in the ratio 4 : 3 and their monthly savings are in the ratio 9 : 5 . If each spends Rs. 3500 per month, find the monthly income of each.

Answer:

Let A’s Income =x \ Rs.

Let B’s Income  = y \ Rs.

\frac{x}{y} = \frac{4}{3}    

\Rightarrow 3x=4y  ... ... ... ... ... ... (i)

\frac{x-3500}{y-3500} = \frac{9}{5}

\Rightarrow 5x-17500=9y-31500

\Rightarrow 5x-9y=-14000 ... ... ... ... ... ... (ii)

Substituting  x= \frac{4}{3} \ y \ in \ ii) 

\Rightarrow 5( \frac{4}{3} y)-9y=-14000

-7y=-14000\times 3  or y=6000 \ Rs.

Hence x= \frac{4}{3} \times 6000=8000 \ Rs.

Therefore :

A’s Income = Rs.\ 8000

B’s Income = Rs.\ 6000

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Question 21: 6 nuts and 5 bolts weigh 278 grams, while 8 nuts and 3 bolts weigh 268 grams. Find the weight of each nut and that of each bolt. How much do 3 nuts and 3 bolts weigh together?

Answer:

Let the Weight of Nut = x \ gm.

Let the Height of Bolt = y \ gm.

 6x+5y=278  ... ... ... ... ... ... (i)

8x+3y=268 ... ... ... ... ... ... (ii)

Solving for x \ and \ y

Multiply i) by 8 and ii) by 6 and Subtract ii) from i)

48x+40y=2224

\underline{(-)48x+18y=1608}

22y=616)   

\Rightarrow y=28

Hence x= \frac{278-5\times 28}{6} =23

Weight of:

Nut = 23 \ gm.

Bolt = 28\ gm.

Therefore 3 Nuts and 3 Bolts will weight = 3\times 23+3\times 28=153 gm.

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Question 22: There are some girls in two classrooms, A and B. If 12 girls are sent from room A to room B, the number of girls in both the rooms will become equal. If 11 girls are sent from room B to room A, then the number of girls in room A would be double the number of girls in room B. How many girls are there in each class room?

Answer:

Let No. of girls in classroom A = x and in B=y

x-12=y+12 

\Rightarrow x-y=24  ... ... ... ... ... ... (i)

2(y-11)=x+11 

\Rightarrow x-2y=-33 ... ... ... ... ... ... (ii)

Solving x \ and \ y ,

Subtract ii) from i)

x-y=24

\underline{(-)  x-2y=-33}

y=57

Hence x=24+57=81

No. of Girls in Class Room:

A=81

B=57

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Question 23: 4 men and 4 boys can do a piece of work in 3 days, while 2 men and 5 boys can finish it in 4 days. How long would it take 1 man alone to do it?

Answer:

Suppose 1 man finished the work in x days and 1 Boy finished the work in y days

There 1 man’s 1 day’s =   \frac{1}{x}

And, 1 Boy’s 1 Day’s work = \frac{1}{y}

Now 4 men and 4 Boys finished the work in 3 days

\Rightarrow \frac{ 4}{x} + \frac{5}{y} = \frac{1}{3}  ... ... ... ... ... ... (i)

Similarly 2 men and 5 boys finished in 4 days

\Rightarrow \frac{ 2}{x} + \frac{5}{y} = \frac{1}{4} ... ... ... ... ... ... (ii)

Solving for x  \ and\  y

Multiply ii) by 2 and Subtract ii) from 2

\Rightarrow \frac{ 4}{x} + \frac{4}{y} = \frac{1}{3}

\Rightarrow \underline{(-) \frac{ 4}{x}+\frac{10}{y}=\frac{1}{2}}

\frac{-6}{y} = \frac{-1}{6}

or y=36

Similarly x= \frac{4}{\frac{1}{3}-\frac{4}{36}} = \frac{4 \times 36}{8} = 18

So one men can finished the work in 18 \ days.

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Question 24: A takes 3 hours longer than B to walk 30 km. But, if A doubles his pace, he is ahead of B by 1 hour 30 minutes. Find the speeds of A and B.

Answer:

Let the Speed of A = x \ km/Hr.

Let the Speed of B = y \ km/Hr.

Time taken By A =   \frac{30}{x}

Time Taken By B = \frac{30}{y} 

\Rightarrow  \frac{30}{x} =\frac{30}{y} +3  ... ... ... ... ... ... (i)

If A Double him Race Time taken by A = \frac{30}{2x} 

Therefore

\frac{30}{2x} + \frac{3}{2} = \frac{30}{y} 

\Rightarrow \underline{(-) \frac{30}{x} = \frac{30}{y}+3}

\frac{-30}{2x} + \frac{3}{2} =-3

\Rightarrow x= \frac{10}{3} =3 \frac{1}{3} Km/Hr.

Hence  \frac{30}{y} = \frac{30}{10} \times 3-3=6  

\Rightarrow y= \frac{30}{6} =5 \ km/Hr.

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Question 25: If the length of a rectangle is reduced by 1 m and breadth increased by 2 m, its area increases by 32 m2. If however, the length is increased by 2 m and breadth reduced by 3 m, then the area is reduced by 49 m2. Find the length and breadth of the rectangle.

Answer:

Let the Length of the rectangle =x

Let the Breadth of the rectangle = y

\Rightarrow (x-1)(y+2)=xy+32

\Rightarrow xy-y+2x-2=xy+32

\Rightarrow 2x-y=34  ... ... ... ... ... ... (i)

\Rightarrow (x+2)(y-3)=xy-49

\Rightarrow 3x-2y-3x-6=xy-49

\Rightarrow 3x-2y=43 ... ... ... ... ... ... (ii)

Solving for x \ and \ y

From i) y=2x-34

Substituting in ii) \Rightarrow 3x-2(2x-34)=43

\Rightarrow 3x-4x+68=43

\Rightarrow  x=25\ m

Hence y=2(25)-34=16\ m

Therefore:

Length = 25\ m

Breadth = 16\ m.